HDU - 3584 Cube (三维树状数组 + 区间修改 + 单点求值)

HDU - 3584 Cube Time Limit: 1000MS   Memory Limit: 65536KB   64bit IO Format: %I64d & %I64u Submit Status Description Given an N*N*N cube A, whose elements are either 0 or 1. A[i, j, k] means the number in the i-th row , j-th column and k-th layer. Initially we have A[i, j, k] = 0 (1 <= i, j, k <= N).  We define two operations, 1: “Not” operation that we change the A[i, j, k]=!A[i, j, k]. that means we change A[i, j, k] from 0->1,or 1->0. (x1<=i<=x2,y1<=j<=y2,z1<=k<=z2).  0: “Query” operation we want to get the value of A[i, j, k].    Input Multi-cases.  First line contains N and M, M lines follow indicating the operation below.  Each operation contains an X, the type of operation. 1: “Not” operation and 0: “Query” operation.  If X is 1, following x1, y1, z1, x2, y2, z2.  If X is 0, following x, y, z.    Output For each query output A[x, y, z] in one line. (1<=n<=100 sum of m <=10000)   Sample Input 2 5 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 2 2 2 0 1 1 1 0 2 2 2   Sample Output 1 0 1   /* Author: 2486 Memory: 5944 KB Time: 202 MS Language: G++ Result: Accepted VJ RunId: 4328542 Real RunId: 14413386 */ //三维的与二维的一样，而二维的与一维的一样 //详解，本博客中解答了题目这么做的原理 #include #include #include #include #include using namespace std; const int MAXN = 100 + 5; int C[MAXN][MAXN][MAXN]; int N, M; int lowbit(int x) { return x & (-x); } void add(int x, int y, int z) { for(int i = x; i <= N; i += lowbit(i)) { for(int j = y; j <= N; j += lowbit(j)) { for(int k = z; k <= N; k += lowbit(k)) { C[i][j][k] ++; } } } } int query(int x, int y, int z) { int ret = 0; for(int i = x; i > 0 ; i -= lowbit(i)) { for(int j = y; j > 0; j -= lowbit(j)) { for(int k = z; k > 0; k -= lowbit(k)) { ret += C[i][j][k]; } } } return ret & 1; } int X, x, y, z, x1, y1, z1, x2, y2, z2; int main() { while(~ scanf("%d%d", &N, &M)) { memset(C, 0, sizeof(C)); while(M --) { scanf("%d", &X); if(X) { scanf("%d%d%d%d%d%d", &x1, &y1, &z1, &x2, &y2, &z2); x2 ++; y2 ++; z2 ++; add(x1, y1, z1); add(x1, y2, z1); add(x1, y1, z2); add(x1, y2, z2); add(x2, y1, z1); add(x2, y2, z1); add(x2, y1, z2); add(x2, y2, z2); } else { scanf("%d%d%d", &x, &y, &z); printf("%d\n", query(x, y, z)); } } } return 0; }